Physics for Scientists and Engineers: A Strategic Approach with Modern Physics, Books a la Carte Edition; Student Workbook for Physics for Scientists ... eText -- ValuePack Access Card (4th Edition)
4th Edition
ISBN: 9780134564234
Author: Randall D. Knight (Professor Emeritus)
Publisher: PEARSON
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Chapter 23, Problem 12CQ
A parallel-plate capacitor consists of two square plates, size
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Suppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rb has charge -Q. The electric field E at a radial distance r from the central axis is given by the function:
E = αe-r/a0 + β/r + b0
where alpha (α), beta (β), a0 and b0 are constants. Find an expression for its capacitance.
First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by:
Calculating the antiderivative or indefinite integral ,
Vab = (-αa0e-r/a0 + β + b0 )
By definition, the capacitance C is related to the charge and potential difference by:
C = /
Evaluating with the upper and lower limits of integration for Vab, then simplifying:
C = Q / ( (e-rb/a0 - e-ra/a0) + β ln() + b0 () )
Suppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rb has charge -Q. The electric field E at a radial distance r from the central axis is given by the function:
E = αe-r/a0 + β/r + b0
where alpha (α), beta (β), a0 and b0 are constants. Find an expression for its capacitance.
Suppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rb has charge -Q. The electric field E at a radial distance r from the central axis is given by the function:
E = αe-r/a0 + β/r + b0
where alpha (α), beta (β), a0 and b0 are constants. Find an expression for its capacitance.
First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by:
Chapter 23 Solutions
Physics for Scientists and Engineers: A Strategic Approach with Modern Physics, Books a la Carte Edition; Student Workbook for Physics for Scientists ... eText -- ValuePack Access Card (4th Edition)
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Ch. 23 - Rank in order, from largest to smallest, the...Ch. 23 - A parallel-plate capacitor consists of two square...Ch. 23 - A small object is released at point 3 in the...Ch. 23 - A proton and an electron are released from rest in...Ch. 23 - Three charges are placed at the comers of the...Ch. 23 - l. What are the strength and direction of the...Ch. 23 - What are the strength and direction of the...Ch. 23 - What are the strength and direction of the...Ch. 23 - What are the strength and direction of the...Ch. 23 - An electric dipole is formed from two charges, q ,...Ch. 23 - An electric dipole is formed from ± 1.0 nC charges...Ch. 23 - An electret is similar to a magnet, but rather...Ch. 23 - The electric field strength 10.0 cm from a very...Ch. 23 - A 10-cm-long thin glass rod uniformly charged to...Ch. 23 - Two 10-cm-long thin glass rods uniformly charged...Ch. 23 - A small glass bead charged to + 6.0 nC is in the...Ch. 23 - The electric field 5.0 cm from a very long charged...Ch. 23 - A 12-cm-long thin rod has the nonuniform charge...Ch. 23 - Two charged rings face each other, 20 cm apart....Ch. 23 - Two 10-cm-diameter charged rings face each other,...Ch. 23 - Two charged disks face each other, 20 cm apart....Ch. 23 - The electric field strength 2.0 cm from the...Ch. 23 - A 20cm20cm cm horizontal metal electrode is...Ch. 23 - Two 2.0-cm-diameter insulating spheres have a 6.0...Ch. 23 - You've hung two very large sheets of plastic...Ch. 23 - A 2.0m X 4.0m flat carpet acquires a uniformly...Ch. 23 - Two circular disks spaced 0.50 mm apart form a...Ch. 23 - A parallel-plate capacitor is formed from two...Ch. 23 - Air "breaks down" when the electric field strength...Ch. 23 - Two parallel plates 1.0 cm apart are equally and...Ch. 23 - a. What is the electric field strength between the...Ch. 23 - Honeybees acquire a charge while flying due to...Ch. 23 - An electron traveling parallel to a uniform...Ch. 23 - The surface charge density on an infinite charged...Ch. 23 - An electron in a vacuum chamber is fired with a...Ch. 23 - A 1.0m -diameter oil droplet (density 900 kg/m3)...Ch. 23 - The permanent electric dipole moment of the water...Ch. 23 - A point charge Q is distance r from a dipole...Ch. 23 - An ammonia molecule (NH3) has a permanent electric...Ch. 23 - What are the strength and direction of the...Ch. 23 - What are the strength and direction of the...Ch. 23 - What are the strength and direction of the...Ch. 23 - Prob. 38EAPCh. 23 - Prob. 39EAPCh. 23 - Derive Equation 23.11 for the field Edipolein the...Ch. 23 - FIGURE P23.41 is a cross section of two infinite...Ch. 23 - FIGURE P23.42 is a cross section of two infinite...Ch. 23 - Prob. 43EAPCh. 23 - Prob. 44EAPCh. 23 - Prob. 45EAPCh. 23 - Prob. 46EAPCh. 23 - Prob. 47EAPCh. 23 - A plastic rod with linear charge density ? is bent...Ch. 23 - An infinite plane of charge with surface charge...Ch. 23 - A sphere of radius R and surface charge density ?...Ch. 23 - Prob. 51EAPCh. 23 - An electron is launched at a 45 angle and a speed...Ch. 23 - The two parallel plates in FIGURE P23.53 are 2.0...Ch. 23 - Prob. 54EAPCh. 23 - Prob. 55EAPCh. 23 - 56. Your physics assignment is to figure out a way...Ch. 23 - Prob. 57EAPCh. 23 - Prob. 58EAPCh. 23 - Prob. 59EAPCh. 23 - Prob. 60EAPCh. 23 - Prob. 61EAPCh. 23 - Prob. 62EAPCh. 23 - In Problems 63 through 66 you are given the...Ch. 23 - Prob. 64EAPCh. 23 - Prob. 65EAPCh. 23 - Prob. 66EAPCh. 23 - A rod of length L lies along the y-axis with its...Ch. 23 - a. An infinitely long sheet of charge of width L...Ch. 23 - a. An infinitely long sheet of charge of width L...Ch. 23 - Prob. 70EAPCh. 23 - Prob. 71EAPCh. 23 - 72. A proton orbits a long charged wire, making ...Ch. 23 - Prob. 73EAP
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- Suppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rb has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = αe-r/a0 + β/r + b0 where alpha (α), beta (β), a0 and b0 are constants. Find an expression for its capacitance. First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: Calculating the antiderivative or indefinite integral , Vab = (-αa0e-r/a0 + β + b0 ) By definition, the capacitance C is related to the charge and potential difference by: C = / Evaluating with the upper and lower limits of integration for Vab, then simplifying: C = Q / ( (e-rb/a0 - e-ra/a0) + β ln() + b0 () )arrow_forwardSuppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rb has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = αe-r/a0 + β/r + b0 where alpha (α), beta (β), a0 and b0 are constants. Find an expression for its capacitance.arrow_forwardSuppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rb has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = αe-r/a0 + β/r + b0 where alpha (α), beta (β), a0 and b0 are constants. Find an expression for its capacitance. First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by:arrow_forward
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